Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{r(5r + 3)}{6} \times \dfrac{-2}{6(5r + 3)} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ r(5r + 3) \times -2 } { 6 \times 6(5r + 3) } $ $ z = \dfrac{-2r(5r + 3)}{36(5r + 3)} $ We can cancel the $5r + 3$ so long as $5r + 3 \neq 0$ Therefore $r \neq -\dfrac{3}{5}$ $z = \dfrac{-2r \cancel{(5r + 3})}{36 \cancel{(5r + 3)}} = -\dfrac{2r}{36} = -\dfrac{r}{18} $